vast A time bubble has been projected into the future to the precise moment of the end of the universe. This is, of course, impossible. --D. Adams, The Hitchhikers Guide to
the Galaxy
There is light at the end of the tunnel. -- proverb The light at the end of the tunnel is just the light of an oncoming train. --R. Lowell
Lecture 14, p 1
Lecture 14: Barrier Penetration and Tunneling
nucleus x U(x) U(x) U0 E AB 0L x 0 C B x
Lecture 14, p 2
A
Act 0
The midterm exam was:
a. b. c. d. e. Just right Too easy Too hard Too long Too long and too hard
Lecture 14, p 3
The Most Difficult Midterm Question
arm 2
laser 8 mW = 700 nm or 500 nm
arm 1 detector
4
c.
Detected Power (mW) 3
2
a. b.
Interference (of single photons and lots of photons) requires two (or more) indistinguishable processes.
1
0 Mirror position
Lecture 14, p 4
Today
Tunneling of quantum particles
Scanning Tunneling Microscope (STM) Nuclear Decay Solar Fusion The Ammonia Maser
The rest of the course:
Next week: 3 dimensions - orbital and spin angular momentum H atom, exclusion principle, periodic table Last week: Molecules and solids. Metals, insulators, semiconductors, superconductors, lasers, . .
Good web site for animations http://www.falstad.com/qm1d/
Lecture 14, p 5
Leaky Particles: Revisited
Due to barrier penetration, the electron density of a metal actually extends outside the surface of the metal! x
Vo Work function EF
Occupied levels
x=0
Assume that the work function (i.e., the energy difference between the most energetic conduction electrons and the potential barrier at the surface) of a certain metal is = 5 eV. Estimate the distance x outside the surface of the metal at which the electron probability density drops to 1/1000 of that just inside the metal.
Lecture 14, p 6
Leaky Particles: Revisited
Due to barrier penetration, the electron density of a metal actually extends outside the surface of the metal! x
Vo Work function EF
Occupied levels
x=0
Assume that the work function (i.e., the energy difference between the most energetic conduction electrons and the potential barrier at the surface) of a certain metal is = 5 eV. Estimate the distance x outside the surface of the metal at which the electron probability density drops to 1/1000 of that just inside the metal.
(x) (0)
K= 2me
2
2 2
= e 2Kx
1 1000
5eV 1.505 eV nm
2
x=
= 11.5 nm-1
1 1 ln 0.3 nm 2 K 1000
(V
0
E ) = 2
2me h
2
= 2
Lecture 14, p 7
Tunneling: Key Points
In quantum mechanics a particle can penetrate into a barrier where it would be classically forbidden.
U(x) U0
I II III
The finite square well:
In region III, E < U0, and (x) has the exponential form D1e-Kx. We did not solve the equations too hard! You did it using the computer in Lab #3.
0
Lx
The probability of finding the particle in the barrier region decreases as e-2Kx.
The finite-width barrier:
Today we consider a related problem a particle approaching a finite-width barrier and tunneling through to the other side. The result is very similar, and again the problem is too hard to solve exactly here: The probability of the particle tunneling through a finite width barrier is approximately proportional to e-2KL where L is the width of the barrier.
Lecture 14, p 8
Tunneling Through a Barrier (1)
U(x) What is the the probability that an incident particle tunnels through the barrier? Its called the Transmission Coefficient, T. Consider a barrier (II) of height U0. U = 0 everywhere else.
Uo I II III
0 0 L x
Getting an exact result requires applying the boundary conditions at x = 0 and x = L, then solving six transcendental equations for six unknowns:
I ( x ) = A1 sin kx + A2 cos kx II ( x ) = B1e Kx + B2e Kx III ( x ) = C1 sin kx + C2 cos kx
A1, A2, B1, B2, C1, and C2 are unknown. K and k are known functions of E. This is more complicated than the infinitely wide barrier, because we cant require that B1 = 0. (Why not?)
Lecture 14, p 9
Tunneling Through a Barrier (2)
In many situations, the barrier width L is much larger than the decay length 1/K of the penetrating wave (KL >> 1). In this case B1 0 (why?), and the result resembles the infinite barrier. The tunneling coefficient simplifies:
T Ge 2KL where G = 16 E E 1 U 0 U0
2m (U 0 E ) 2
U(x) U0 E 0L
x
K=
This is nearly the same result as in the leaky particle example! Except for G: We will often ignore G. (Well tell you when to do this.) The important result is e-2KL.
4 3
G
2 1 0
0 0.25 0.5 0.75 1
E/U0
Lecture 14, p 10
Act 1
Consider a particle tunneling through a barrier. 1. Which of the following will increase the likelihood of tunneling? a. decrease the height of the barrier b. decrease the width of the barrier c. decrease the mass of the particle
0L U0 E U(x)
x
2. What is the energy of the emerging particles? a. < initial energy b. = initial energy c. > initial energy
Lecture 14, p 11
Solution
Consider a particle tunneling through a barrier. 1. Which of the following will increase the likelihood of tunneling? a. decrease the height of the barrier b. decrease the width of the barrier c. decrease the mass of the particle
0L U0 E U(x)
x
T e 2KL
Decreasing U0 or me will decrease K.
2. What is the energy of the emerging particles? a. < initial energy b. = initial energy c. > initial energy
Lecture 14, p 12
Solution
Consider a particle tunneling through a barrier. 1. Which of the following will increase the likelihood of tunneling? a. decrease the height of the barrier b. decrease the width of the barrier c. decrease the mass of the particle
0L U0 E U(x)
x
T e 2KL
Decreasing U0 or me will decrease K.
2. What is the energy of the emerging particles? a. < initial energy b. = initial energy c. > initial energy Probability Energy
Lecture 14, p 13
The barrier does not absorb energy from the particle. The amplitude of the outgoing wave is smaller, but the wavelength is the same. E is the same everywhere.
Nature 434, 361 - 364 (17 March 2005) Current measurement by real-time counting of single electrons
JONAS BYLANDER, TIM DUTY & PER DELSING
Example: Electrons in Example: Nanoscale devices Nanoscale
~40 fA
~80 fA
~120 fA
Electrons that successfully tunnel through the 50 junctions are detected using a fast single-electron transistor (SET).
Application: Tunneling Microscopy
One can use barrier penetration to measure the electron density on a surface. x
Metal tip
Real STM tip
Scanning Tunneling Microscope images
Na atoms on metal: DNA Double Helix:
material
STM tip ~ 1 nm
material STM tip
STM demo: http://www.quantum-physics.polytechnique.fr/en/
Barrier penetration is a wave phenomenon, not only QM. It is used in optical microscopes also. See:
http://en.wikipedia.org/wiki/Total_internal_reflection_fluorescence_microscope Lecture 14, p 15
The STM
The STM (scanning tunneling microscope) tip is L = 0.18 nm from a metal surface. An electron with energy of E = 6 eV in the metal approaches the surface. Assume the metal/tip gap is a potential barrier with a height of Uo = 12 eV. What is the probability that the electron will tunnel through the barrier? U(x) U0 E
metal STM tip
0L
gap
x
Lecture p 14, 16
The STM
The STM (scanning tunneling microscope) tip is L = 0.18 nm from a metal surface. An electron with energy of E = 6 eV in the metal approaches the surface. Assume the metal/tip gap is a potential barrier with a height of Uo = 12 eV. What is the probability that the electron will tunnel through the barrier? U(x) U0 E
metal STM tip
0L
gap
x
T Ge 2KL = 4 e 2(12.6)(0.18) = 4(0.011) = 4.3%
G = 16
E E 1 1 1 = 16 1 = 4 U0 U0 2 2
K=
2me 2m (U0 E ) = 2 2 e (U0 E ) 2 h 6 eV 12.6 nm1 2 1.505 eV-nm
T << 1, so our use of the KL >> 1 approximation is justified.
= 2
Q: What will T be if we double the width of the gap?
Lecture 14, p 17
Tunneling and Radioactivity
In large atoms (e.g., Uranium), the nucleus can be unstable to the emission of an alpha particle (a He nucleus). This form of radioactivity is a tunneling process, involving transmission of the alpha particle from a low-energy valley through a barrier to a lower energy outside.
U(x)
interior of nucleus
outside
outside
A
B
C
B
A
Why do we observe exponential decay?
leaks out from C through B to A the particle tunnels out. The leakage is slow (T << 1), so just outside the barrier stays negligible. The shape of remaining in B-C shows almost no change: Its amplitude slowly decreases. That is, Pinside is no longer 1. The rate at which probability flows out is proportional to Pinside (by linearity) exponential decay in time.
dx = Ax dt
x = e At = e t /
t1/2 = ( ln 2) is the half life of the substance
Lecture 14, p 18
-Radiation: Illustrations of the enormous Radiation: range of decay rates in different nuclei
Consider a very simple model of a-radiation:
Assume the alpha particle (m = 6.64 x 10-27 kg) is trapped in a nucleus which presents a square barrier of width L and height Uo. To find the decay rate we consider: (1) the attempt rate at which the alpha particle of energy E inside the nucleus hits the barrier
Rough estimate with E ~ 5 to 10 MeV: the alpha particle makes about 1021 attempts per second (~velocity/nuclear diameter)
(2) the tunneling probability for an alpha particle with energy E each time the particle hits the barrier. [For this order of magnitude calculation you may neglect G.] Here we use
T e
2 KL
K=
2m U E) 2( 0
Because of the exponential this factor can vary enormously!
Act 3
Polonium has an effective barrier width of ~10 fermi, leading to a tunneling probability of ~10-15. Now consider Uranium, which has a similar barrier height, but an effective width of about ~20 fermi. Estimate the tunneling probability in Uranium: a. 10-30 b. 10-14 c. 10-7
Solution
Polonium has an effective barrier width of ~10 fermi, leading to a tunneling probability of ~10-15. Now consider Uranium, which has a similar barrier height, but an effective width of about ~20 fermi. Estimate the tunneling probability in Uranium: a. 10-30 b. 10-14 c. 10-7
Think of it this way there is a 10-15 chance to get through the first half of the barrier, and a 10-15 chance to then get through the second half. Alternatively, when we double L in 2 KL
T e
this is equivalent to squaring the transmission T. Polonium: Using 1021 attempts at the barrier per second, the probability of escape is about 106 per second decay time ~1 s. Uranium: Actually has a somewhat higher barrier too, leading to P(tunnel) ~ 10-40 decay time ~1010 years!
Tunneling Example: The Sun
The solar nuclear fusion process starts when two protons fuse together. In order for this reaction to proceed, the protons must touch (approach to within 10-15 m of each other). The potential energy, U(r), looks something like this: The temperature of the suns core is T ~ 1.3x107 K. This corresponds to an average kinetic energy: kBT = 2 x 10-16 J (kB = Boltzmans constant) At r = 10-15 m the height of the Coulomb barrier is: U(r) = (1/40)e2/r = (9x109)x(1.6x10-19 C)2/10-15 m = 2 x 10-13 J Thus, the protons in the sun very rarely have enough thermal energy to go over the Coulomb barrier. How do they fuse then? By tunneling through the barrier!
Lecture 14, p 22
U(r)
Coulomb repulsion
kBT 0
10-15 m Nuclear binding
r
Another Consequence of Tunneling
Consider a situation in which a particle (e.g., an electron or an atom) can be in either of two wells separated by a potential barrier. Is the particle on the left or right? Both! If the barrier is finite, the wave function extends into both wells
d
Lowest energy state: is small but non-zero inside the barrier.
d
E1
Here is the state with the next higher energy: Why does this state have higher energy?
Note that the potential is symmetric about the middle of the barrier. Therefore, the energy states must be either symmetric or antisymmetric. Also, remember that there are n-1 nodes.
E2
Lecture 14, p 23
Energy Splitting in a Double Well
Suppose the particle starts out in the left well. What is the time dependence of the probability? From the graphs of , we can see that, initially, = 1 + 2 (to get cancellation on the right). As discussed last lecture, the particle oscillates between the wells with an oscillation period, T = h/(E2 - E1). Therefore, E = E2E1 depends on the tunneling rate. A double well with a high or wide barrier will have a smaller E than one with a low or narrow barrier. Also, E will become larger as the energy increases (i.e., as U0 - E decreases).
d
E2 E1
Lecture 14, p 24
Act 4
You are trying to make a laser that emits violet light ( = 400 nm), based on the transition an electron makes U0 between the ground and first-excited state of a double quantum well as shown. Your first sample emitted at E1 = 390 nm. What could you modify to shift the wavelength to 400 nm? a. decrease the height of the barrier b. increase the height of the barrier c. decrease the width of the barrier
E2
Lecture 14, p 25
Solution
You are trying to make a laser that emits violet light ( = 400 nm), based on the transition an electron makes U0 between the ground and first-excited state of a double quantum well as shown. Your first sample emitted at E1 = 390 nm. What could you modify to shift the wavelength to 400 nm? a. decrease the height of the barrier b. increase the height of the barrier c. decrease the width of the barrier The frequency of the electron oscillating between the left and right well was too high the probability to tunnel was too high! You can reduce this by increasing the barrier height. The wavelength of the emitted photon was too low the frequency of the photon was too high the energy splitting between the ground and first-excited state was too large. Raising the barrier makes the difference in energy E2-E1 smaller. Why?
Lecture 14, p 26
E2
Solution - More
As we raise the height of the central barrier, the coupling between the two wells decreases. In the limit of an infinite barrier, it looks like two independent wells same wavefunction curvature for both the symmetric (ground state) and anti-symmetric (1st excited state) wavefunctions same kinetic energy, i.e., degenerate solutions.
0 1
0
1
Next Week
3-Dimensional Potential Well Schrdingers Equation for the Hydrogen Atom
Lecture 14, p 28